package com.ljc;

import com.ljc.dto.TreeNode;

/**
 * @author clj
 * @date 2022/12/2
 * @desc 判断对称二叉树
 * Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).
 *
 * Example 1:
 * Input: root = [1,2,2,3,4,4,3]
 * Output: true
 *
 * Example 2:
 * Input: root = [1,2,2,null,3,null,3]
 * Output: false
 */
public class E101SymmetricTree {
    public static void main(String[] args) {
        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(2);
        root.right = new TreeNode(2);

//        root.left.left = new TreeNode(4);
        root.left.right = new TreeNode(5);

        root.right.left = new TreeNode(5);
//        root.right.right = new TreeNode(4);

        System.out.println(isSymmetric(root));
    }

    public static boolean isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        //调用递归函数，比较左节点，右节点
        return dfs(root.left, root.right);
    }

    /**
     * 深度优先搜索算法（英语：Depth-First-Search，DFS）是一种用于遍历或搜索树或图的算法
     * @param left
     * @param right
     * @return
     */
    public static boolean dfs(TreeNode left, TreeNode right) {
        //递归的终止条件是两个节点都为空
        //或者两个节点中有一个为空
        //或者两个节点的值不相等
        if (left == null && right == null) {
            return true;
        }
        if (left == null || right == null) {
            return false;
        }
        if (left.val != right.val) {
            return false;
        }
        //再递归的比较 左节点的左孩子 和 右节点的右孩子
        //以及比较  左节点的右孩子 和 右节点的左孩子
        return dfs(left.left, right.right) && dfs(left.right, right.left);
    }


}
